10 10 UTG
 

I would like to know what is the probability and how to calculate similar schenarios where the probability that of 9 random hands that if you have 10 10 UTG that one of these 9 hands is JJ QQ KK or AA.


Can someone suggest a computer program or book that explains or has tables to figure out situations like these.


Has anyone used Hold'Em Analyzer or another such program.

Re: 10 10 UTG
 

It's really not that tricky. There are 52!/(2!*(52-2)!) = 1326 number of different holdem hands. If you hold AA you can have it 6 different ways. The same goes for any pair. So, AA, KK, QQ, JJ, that sums up to 24 ways.


24 out of 1326 for a probability of 1.81%


Now, instead of calculating the probability that one or more players have such a pair we will calculate the probability that none of them has one. The probability that one player does not hold a pair is 98.19% so the probability that none of the remaining 9 players have one is:


0.9819^9 = 0.8484


So, there is about an 85% chance that the players do not have such a pair. Consequently there is about a 15% chance that one of the players has such a pair.


Hope this didn't cause as many questions as it answered. [img]/images/wink.gif[/img]


Greets, Andreas

Re: 10 10 UTG
 

Actually it is 16.3%.

Remember, two cards are already gone.


But yes, very easy to compute.


Pretty interesting you are that far behind 1/6 of the time.

You\'re both wrong
 

It's very easy to compute the wrong answer, and a relative pain in the butt to compute the correct one:


1-

[(50*49/2)-24]/(50*49/2)*

[(48*47/2)-24]/(48*47/2)*

[(46*45/2)-24]/(46*45/2)*

[(44*43/2)-24]/(44*43/2)*

[(42*41/2)-24]/(42*41/2)*

[(40*39/2)-24]/(40*39/2)*

[(38*37/2)-24]/(38*37/2)*

[(36*35/2)-24]/(36*35/2)*

[(34*33/2)-24]/(34*33/2)


= 22%.


The denominator will come up in other problems with 9 opponents, so it is worth remembering its value(2.2575x10^26).


Note that the odds of a given player having one of these hands is 24/1326=1.8%, but if the first 9 players do not have one of these hands, the chance that the final player has it is 4.3% or more than twice as great.

make that 23.4% *NM*
 

Re: You\'re both wrong
 

Please explain which factorials you were using for each of the 9 hands and why you multiplied them instead of adding them.


Also have you considered the cards UTG + 1 has could be anything including AJ and 33 one of them depletes two of the cards that could make the possible 4 hands the other does not. Please explain your numbers.

Re: You\'re both wrong
 

I am multiplying the probability that the first player doesn't have one of the 24 hands, times the probability that the second player doesn't have it, etc. The denominator in each case is the total number of hands each player can make out of the remaining cards. The numerator is this number minus the 24 hands.


You are correct that depending on which cards each player receives, some of the 24 hands may no longer be possible. I am not taking this into account. That would make the problem even more difficult, and the result would be lower than I have reported.

Better Answer
 

Let's try a different approach which should account for everything:


The total number of ways to deal 50 cards to 9 players (18 cards total) is:


50*49*48*47*46*45*...*33/2


We divide by 2 to avoid counting every player's possible hand twice.


The number of ways for exactly 1 player to get JJ-AA is:


24*9*48*47*46*45*...*33/2


Since there are 24 hands and 9 players that can get them. So the probability of exactly 1 player having JJ-AA is (24*9)/(50*49) = 8.8%.


The number of ways exactly 2 players can have JJ-AA is:


C(24,2)*C(9,2)*46*45*...*33/2


Since there are C(24,2) = 24*23/2 hands and

C(9,2) = 9*8/2 players that can get them.


So the probability of exactly 2 players having JJ-AA is C(24,2)*C(9,2)/(50*49*48*47) = 1.8%


We could go on like this computing probailities for exactly 3,4,5...players, but the rest of the terms become too small to worry about. So the total probability is 8.8% + 1.8% = 10.6%.


So in summary, the original posters ignored the effect that each players cards had on the others. My original post accounted for the fact that each successive player has fewer cards to choose from which made it more likely to draw one of the combinations, but it failed to take into account an even bigger effect which is that the exact cards drawn tend to block certain hands. The net result was that I moved the percentage about the right amount in the wrong direction. It should actually go down to 10.6% in light of the above.

Re: Better Answer
 

I'm not sure what you mean by "number of ways to deal 50 cards to 9 players". If we don't care about the order of the cards, but do care about the order of the players, then the number of ways to do this is P(50,18)/(2^9) = 50*49*...*33/(2^9). This is because each of the 9 hands can be ordered in 2 ways.

Re: Better Answer
 

Another problem here is how to come up with the number of ways of assigning 2 hands, given you've picked the 2 players. It's not C(24,2), because once you assign one of the players a hand, Qh Qd, say, then you've eliminated hands like Qh Qs, or Qd Qc. So, C(24,2) counts Qh Qs + Qh Qd, whereas this possibility is impossible in the problem.